Chapter 10, Problem 10.5P

Two voltage waves of equal amplitude V₀ and radian frequency $\omega$ propagate in the forward z direction in a lossy transmission line having attenuation coefficient $\alpha$, and characteristic impedance $Z_0=\left|Z_0\right|e^{i\delta }$. One wave is shifted from the other by $\varphi$ radians. (a) Find an expression for the net voltage wave formed by the superposition of the two voltages. Your result should be a single wave function in real instantaneous form. (b) Find an expression for the net current in the line, again in the form of a single wave function. (c) Find an expression for the average power in the line.

To determine

(a)

An expression for the net voltage wave which is formed by the superposition of the two voltages.

Answer to Problem 10.5P

The expression for the net voltage wave formed by the superposition of the two voltages is given by $V(Z)=2V_0{e}^{-\alpha Z}{e}^{-J(\beta Z-\frac{\varphi }{2})}\cdot \cos \left(\frac{\varphi }{2}\right)$ . In instantaneous form, the equation is given as $V(Z,t)=2V_0{e}^{-\alpha Z}\cdot \cos \left(\frac{\varphi }{2}\right)\cdot \cos \left(\omega t-\beta Z+\frac{\varphi }{2}\right)$.

Explanation of Solution

Given information:

The amplitude of two voltage waves is equal to $V_0$ and radian frequency $\omega$ . They propagate in a lossy transmission line in the forward $z$ direction whose attenuation coefficient is $\alpha$ , and characteristic impedance is $Z_0=\left|Z_0\right|e^{j\delta }$ . Also, the phase shift between the two waves is $\Phi$ radians.

Calculation:

We write the equation of voltage when propagating in the forward $z$ direction in a lossy transmission line having attenuation coefficient $\alpha$ to determine the expression for net voltage.

   $\begin{array}{c}V(Z)=V_0{e}^{-\alpha Z}{e}^{-J\beta Z}[1+e^{J\varphi }]=V_0{e}^{-\alpha Z}{e}^{-J\beta Z}[1+\cos \varphi +J\sin \varphi ]\\ =V_0{e}^{-\alpha Z}{e}^{-J\beta Z}\sqrt{[{\left(1+\cos \varphi \right)}^2+{\sin }^2\varphi ]}{e}^{J\theta }\\ \text{where, }\theta =\frac{\varphi }{2}\\ =2V_0{e}^{-\alpha Z}{e}^{-J\beta Z}\cdot \cos (\frac{\varphi }{2})\cdot e^{J\frac{\varphi }{2}}=2V_0{e}^{-\alpha Z}{e}^{-J(\beta Z-\frac{\varphi }{2})}\cdot \cos (\frac{\varphi }{2})\end{array}$

Hence, the expression is given by $V(Z)=2V_0{e}^{-\alpha Z}{e}^{-J(\beta Z-\frac{\varphi }{2})}\cdot \cos (\frac{\varphi }{2})$.

In time domain,

   $V(Z,t)=2V_0{e}^{-\alpha Z}\cdot \cos (\frac{\varphi }{2})\cdot \cos (\omega t-\beta Z+\frac{\varphi }{2})$.

Conclusion:

The expression for the net voltage wave formed by the superposition of the two voltages is given by $V(Z)=2V_0{e}^{-\alpha Z}{e}^{-J(\beta Z-\frac{\varphi }{2})}\cdot \cos (\frac{\varphi }{2})$ . In instantaneous form, the equation is given as $V(Z,t)=2V_0{e}^{-\alpha Z}.\cos (\frac{\varphi }{2}).\cos (\omega t-\beta Z+\frac{\varphi }{2})$.

To determine

(b)

An expression for the net current in the line.

Answer to Problem 10.5P

The expression for the net current in the line is

$I(Z)=\frac{2V_0{e}^{-\alpha Z}}{Z_0}\cos (\frac{\varphi }{2})\cdot e^{-J(\beta Z+\delta -\frac{\varphi }{2})}$.

Explanation of Solution

Given information:

The amplitude of two voltage waves is equal to $V_0$ and radian frequency $\omega$ . They propagate in a lossytransmission line in the forward $z$ direction whose attenuation coefficient is $\alpha$ , and characteristic impedance is $Z_0=\left|Z_0\right|e^{j\delta }$ . Also, the phase shift between the two waves is $\Phi$ radians.

Calculation:

Use the previously obtained equation to determine the expression for net current in the line.

We previously obtained

   $V(Z)=2V_0{e}^{-\alpha Z}{e}^{-J\left(\beta Z-\frac{\varphi }{2}\right)}.\cos \left(\frac{\varphi }{2}\right)$

(1) Therefore, the net current in the line is given by

$$

$I(Z)=\frac{V(Z)}{Z_0}=\frac{V(Z)}{Z_0{e}^{J\delta }}$

Using the equation (1) for voltage here, we get

$I(Z)=\frac{2V_0{e}^{-\alpha Z}}{Z_0}\cos (\frac{\varphi }{2})\cdot e^{-J(\beta Z+\delta -\frac{\varphi }{2})}$

In time domain,

   $I(Z,t)=\frac{2V_0}{Z_0}{e}^{-\alpha Z}\cdot \cos (\frac{\varphi }{2})\cdot \cos (\omega t-\beta Z+\frac{\varphi }{2}-\delta )$

Conclusion:

The expression for the net current in the line is $I(Z)=\frac{2V_0{e}^{-\alpha Z}}{Z_0}\cos (\frac{\varphi }{2})\cdot e^{-J(\beta Z+\delta -\frac{\varphi }{2})}$ .In instantaneous form, the equation is given as $I(Z,t)=\frac{2V_0}{Z_0}{e}^{-\alpha Z}\cdot \cos (\frac{\varphi }{2})\cdot \cos (\omega t-\beta Z+\frac{\varphi }{2}-\delta )$.

To determine

(c)

The expression for the average power in the line.

Answer to Problem 10.5P

The expression for the average power in the line is $P=\frac{2V_0{}^2}{Z_0}\left[{\cos }^2\left(\frac{\varphi }{2}\right)\cdot e^{-2\alpha Z}\cdot \cos \delta \right]\text{W}$

Explanation of Solution

Given information:

The amplitude of two voltage waves is equal to $V_0$ and radian frequency $\omega$ . They propagate in a lossytransmission line in the forward $z$ direction whose attenuation coefficient is $\alpha$ , and characteristic impedance is $Z_0=\left|Z_0\right|e^{j\delta }$ . Also, the phase shift between the two waves is $\Phi$ radians.

Calculation:

We know of the power flow expression $P=\frac{1}{2}\mathrm{Re}[V(Z).I*(Z)].$

We use this expressionto determine the expression for the average power in the line.

Power flow is given by,

   $\begin{array}{c}P=\frac{1}{2}\mathrm{Re}[V(Z)\cdot I*(Z)]\\ =\frac{1}{2Z_0}{[}^2\cdot [\cos \delta ]\\ =\frac{1}{2Z_0}[4V_0{}^2\cdot {\cos }^2(\frac{\varphi }{2})\cdot e^{-2\alpha Z}\cdot \cos \delta ]\\ P=\frac{2V_0{}^2}{Z_0}\left[{\cos }^2\left(\frac{\varphi }{2}\right)\cdot e^{-2\alpha Z}\cdot \cos \delta \right]\text{W}\end{array}$

Where, * represents the complex conjugate and Re represents the real part.

Conclusion:

The expression for the average power in the line is $P=\frac{2V_0{}^2}{Z_0}\left[{\cos }^2\left(\frac{\varphi }{2}\right)\cdot e^{-2\alpha Z}\cdot \cos \delta \right]\text{W}$